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Let $L$ be a number field. Is it possible to define its ring of integers $R$ by saying it's the subring with (in a fuzzy sense) the "most" irreducibles?

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It seems you're looking for the universal property of normalization, which is the geometric version of "taking the integral closure". See… You might also want to replace "irreducibles" by "prime ideals" in your question. – François Brunault Feb 14 '12 at 13:27
You could make "most" less fuzzy by asking: Does the ring of integers satisfy the property that if there is a subring $S$ of $L$ in which $x\in S$ is irreducible, then $x\in R$ and is irreducible or a product of irreducibles. One problem is that, unless the class number is 1, the ring of integers is not a UFD. So irreducibles behave badly. This is why, historically, one uses prime ideals rather than irreducibles. We have unique factorization of prime ideals. – Pace Nielsen Feb 14 '12 at 22:03

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